How do you factor completely #9t^3+18t-t^2-2#?
1 Answer
Apr 24, 2016
#=(9t-1)(t^2+2)#
#=(9t-1)(t-sqrt(2)i)(t+sqrt(2)i)#
Explanation:
Normally I would rearrange in standard form first, but we can factor by grouping as it is, so...
#9t^3+18t-t^2-2#
#=(9t^3+18t)-(t^2+2)#
#=9t(t^2+2)-1(t^2+2)#
#=(9t-1)(t^2+2)#
The remaining quadratic factor can be factored using Complex coefficients...
#=(9t-1)(t-sqrt(2)i)(t+sqrt(2)i)#